There is a multi-apartment building with $3$ stories and $4$ apartments at each story. In each apartment lives one person. Three random inhabitants of this building are standing outside the building. What the probability that each of them live on a separate floor (event B).
I want to solve this problem using combinatorics approach. The answer that I have in my book is:
$$P(B) = \frac{|B|}{|\Omega|} = \frac{12 \cdot 8 \cdot 4}{12 \cdot 11 \cdot 10}$$
To best of my understanding the logic goes as follows:
(a) total number of possibilities: ordered sample ($3$ out of $12$)
(b) needed possibilities: first we take a person from any apartment ($12$ possibilities), then a person from two remaining floors ($8$ possibilities) and finally a person from one remaining floor ($4$ possibilities).
My question: When I choose people in this problem I do not think we care about the order. So, I think the sample should be unordered. Probably mathematically it does not matter because the “order” factor is in both numerator and denominator is the same. But if indeed I solve the problem as the unordered sample, how I calculate the possibilities in numerator ($|B|$)?
Many thanks.
Your understanding of the solution is correct.
To do it without taking the order of selection into account, observe that there are $\binom{12}{3}$ ways to select three of the twelve apartments. The favorable cases are those in which one of the four apartments on each floor is occupied by the inhabitants standing outside the building. Hence, the probability that each of the three inhabitants of the the three-story building lives on a different floor is $$\dfrac{\dbinom{4}{1}\dbinom{4}{1}\dbinom{4}{1}}{\dbinom{12}{3}}$$
You should verify that this gives the same probability as the solution stated in the book.